In a two-dimensional problem, motion happens in the \(xy\)-plane. The angular momentum vector points along the
\(z\)-axis, either out of the page or into the page.
\[
\vec L=\vec r\times \vec p.
\]
1. Particle angular momentum in 2D
For a particle of mass \(m\),
\[
\vec p=m\vec v.
\]
If the particle is at position
\[
\vec r=(x,y),
\]
and has velocity
\[
\vec v=(v_x,v_y),
\]
then the \(z\)-component of angular momentum is
\[
L_z=m(xv_y-yv_x).
\]
If the pivot is not the origin, use
\[
r_x=x-x_0,
\qquad
r_y=y-y_0.
\]
Then
\[
L_z=m(r_xv_y-r_yv_x).
\]
2. Direction of \(L_z\)
The sign of \(L_z\) tells the direction of angular momentum:
- \(L_z>0\): angular momentum points out of the page, counterclockwise sense.
- \(L_z<0\): angular momentum points into the page, clockwise sense.
- \(L_z=0\): no angular momentum about the chosen pivot.
This is the same right-hand-rule direction used for cross products.
3. Multiple particles
For a system of particles, total angular momentum is the sum of all particle contributions:
\[
L_{z,\mathrm{particles}}
=
\sum_i m_i(r_{xi}v_{yi}-r_{yi}v_{xi}).
\]
Positive and negative contributions can partially cancel.
4. Rigid-body angular momentum in 2D
A disk or rigid body rotating about the same \(z\)-axis has angular momentum
\[
L_{\mathrm{disk}}=I\omega.
\]
Here, \(I\) is the moment of inertia about the pivot axis and \(\omega\) is angular speed.
5. Total angular momentum
If particles and a rotating disk are both present, the total angular momentum is
\[
L_{z,\mathrm{total}}
=
\sum_i m_i(r_{xi}v_{yi}-r_{yi}v_{xi})
+
I\omega.
\]
6. Angular momentum conservation in collisions
During a short collision, angular momentum about a fixed pivot is conserved if the external torque impulse about
that pivot is negligible:
\[
L_{z,i}=L_{z,f}.
\]
This is why angular momentum is useful for particle–disk impacts, grazing collisions, and rotational capture.
7. Particle sticks to a disk
Suppose a particle hits a rotating disk and sticks at distance \(r\) from the pivot. The incoming angular momentum is
\[
L_i=I\omega_i+m(r_xv_y-r_yv_x).
\]
After sticking, the particle rotates with the disk. The final moment of inertia becomes
\[
I_f=I+mr^2.
\]
Conservation of angular momentum gives
\[
I\omega_i+m(r_xv_y-r_yv_x)
=
(I+mr^2)\omega_f.
\]
Therefore,
\[
\omega_f=
\frac{I\omega_i+m(r_xv_y-r_yv_x)}
{I+mr^2}.
\]
8. Outgoing particle and disk spin transfer
If a particle leaves with final position and velocity, its final angular momentum is
\[
L_{p,f}=m(r_{xf}v_{yf}-r_{yf}v_{xf}).
\]
If the disk final angular speed is unknown, conservation gives
\[
L_i=L_{p,f}+I\omega_f.
\]
Solving for the final disk speed:
\[
\omega_f=\frac{L_i-L_{p,f}}{I}.
\]
9. Energy is not always conserved
Angular momentum conservation does not automatically mean kinetic energy conservation. In an inelastic collision,
kinetic energy can become heat, sound, deformation, or internal energy.
Translational kinetic energy is
\[
K_{\mathrm{particle}}=\frac12mv^2.
\]
Rotational kinetic energy is
\[
K_{\mathrm{rot}}=\frac12I\omega^2.
\]
Comparing \(K_i\) and \(K_f\) helps identify whether a collision is elastic, inelastic, or requires external work.
10. Worked example: particle grazing a rotating disk
A particle has
\[
m=0.20\ \mathrm{kg},
\qquad
\vec r=(-1.0,\ 0.40)\ \mathrm{m},
\qquad
\vec v=(5.0,\ 0)\ \mathrm{m/s}.
\]
Its angular momentum is
\[
L_p=m(xv_y-yv_x).
\]
\[
L_p=(0.20)[(-1.0)(0)-(0.40)(5.0)].
\]
\[
L_p=-0.400\ \mathrm{kg\,m^2/s}.
\]
If the disk has
\[
I=0.30\ \mathrm{kg\,m^2},
\qquad
\omega_i=1.0\ \mathrm{rad/s},
\]
then
\[
L_{\mathrm{disk},i}=I\omega_i=(0.30)(1.0)=0.300\ \mathrm{kg\,m^2/s}.
\]
Total initial angular momentum:
\[
L_i=-0.400+0.300=-0.100\ \mathrm{kg\,m^2/s}.
\]
The negative sign means the net angular momentum points into the page.
Key idea: in 2D, angular momentum is a signed scalar \(L_z\), but it still represents a vector direction along the
\(z\)-axis.
